How many ways to put 100 distinguishable particles into 6 boxes?

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Homework Help Overview

The problem involves determining the number of ways to place 100 distinguishable gas particles into 6 adsorbing locations on a filter surface, where each location can hold only one particle. The context suggests a combinatorial approach to the distribution of particles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the problem statement, particularly the distinction between arrangements and combinations. Some explore the factorial approach for arrangements, while others suggest a combinatorial interpretation based on subsets.

Discussion Status

The discussion is ongoing, with differing interpretations of the problem's requirements. Some participants have proposed calculations based on combinations, while others maintain that the arrangement of particles is significant. There is no explicit consensus on the correct interpretation or method.

Contextual Notes

Participants note the constraint that each box can only accommodate one particle, which influences their reasoning about the problem. The use of terms like "how many ways" is also debated in terms of its implications for the solution approach.

bkraabel
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Homework Statement


An adsorbing filter allows gas particles to stick to locations
on the filter surface. Once a particle sticks to a location, that
location is filled. The filter can no longer remove gas particles
when all locations are filled. Each 1.0 nm 2 of the filter surface
has six adsorbing locations, each capable of adsorbing one gas
particle. In the volume just adjacent to one 1.0 nm 2 area, there
are 100 gas particles. Each of these particles has a slightly dif-
ferent energy, making each particle unique. In how many ways
can these 100 particles adsorb onto the six locations?


Homework Equations





The Attempt at a Solution


Consider first 6 particles. You can put the first particle in one of 6 boxes, the second particle in one of 5 remaining boxes, etc., which gives 6! ways to put 6 particles into 6 boxes.

Consider 7 particles. The excluded particle can be thought of as being in the "excluded" box, so you get 7! ways to put 7 distinguishable particles into 6 boxes.

Consider 8 particles. You still have 6! ways to put 6 particles in the 6 boxes. Multiply this by the number of ways can you can put 2 particles out of 8 into the "excluded" box. This latter number is 8 X 7, so the total number of ways to put 8 distinguishable particles into 6 boxes is
6! X 7 X 8 = 8!.

For 9 particles, I think you can have 7 X 8 X 9 ways to put 3 distinguishable particles out of 9 into the "excluded" box, so the total number of ways to put 9 distinguishable particles into 6 boxes is
6! X 7 X 8 X9 = 9!

Following this logic, the total number of ways to put 100 distinguishable particles into 6 boxes is 100!.

Is this correct? Or am I missing something?
thanks
 
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The problem statement says that each box can accommodate only one particle. This problem requires you to find the number of combinations of 100 items taken 6 at a time.
 
Not to sound redundant, but the key word here is: combinations
 
So if I get your drift, then the number of 6-element subsets (where the order of the 6 elements doesn't matter) in a set of 100 distinguishable elements would be 100 choose 6, which is 1.2 X 10^9. Is that the idea?
thanks
 
bkraabel said:
In how many ways can these 100 particles adsorb onto the six locations?

bkraabel said:
So if I get your drift, then the number of 6-element subsets (where the order of the 6 elements doesn't matter) in a set of 100 distinguishable elements would be 100 choose 6, which is 1.2 X 10^9. Is that the idea?
thanks

I have a different on the question. For me, the use of "how many ways" implies that it matters which particle is where on the filter.
 
bkraabel said:
So if I get your drift, then the number of 6-element subsets (where the order of the 6 elements doesn't matter) in a set of 100 distinguishable elements would be 100 choose 6, which is 1.2 X 10^9. Is that the idea?
thanks
Yes. This is the correct answer.
 
Chestermiller said:
Yes. This is the correct answer.

As I said above, I disagree. I don't think this is what the problem is asking.
 

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